Lab #: Electrical Measurements I esistance Goal: Learn to measure basic electrical quantities; study the effect of measurement apparatus on the quantities being measured by investigating the internal resistances of digital multimeters (DMMs) and batteries. Learn the concept of equivalent circuits. Equipment: Battery, DMMs, assorted resistors and a variable resistor box, Proto-Board, connectors, jumper wires,. Internal esistance of a oltmeter When we want to know the voltage between two points, we connect a voltmeter (a DMM set to a voltage scale, see Figure ) between the points and read the result. Images of Typical Digital Multimeters (DMMs) Select Function (olts, Ohms, Amps) with Dial. Hit ange button repeatedly to select range (e.g. 0m,, 0, 00). Make sure not set to autorange. Select Function (olts, Ohms, Amps) and ange (e.g. 0m,, 0, 00) with Dial. Figure : For both types of DMM, make sure you correctly set the Function dial to DC voltage (symbol or ) for measurements of steady voltages and to AC voltage (symbol ) for measurements of oscillating voltages. Always make sure that one of your connections is to Common (also called Ground) and the other to the connection appropriate for the voltage, current, or resistance measurements you are making (different on different DMMs).
An ideal multimeter has infinite internal resistance and has no effect on the voltages in the circuit we are studying, but any real voltmeter has a finite resistance which changes the voltage we are trying to measure. A real voltmeter always draws some current and thus represents a (small) load on the circuit. We can think of a real voltmeter as if it were an equivalent circuit with finite resistance in parallel with an ideal voltmeter, as shown in Figure 2. In this exercise we will find this internal resistance so you can estimate when it might have a significant effect on the voltage measurements you will make later in the course. In order to measure the internal resistance of your voltmeter, set up a circuit with two DMMs as shown in Figures 2 and 3. Why is not equal to 2? This circuit is a voltage divider; with series resistor SS. Show that 2 = SS +. Solve for in terms of, 2 and SS The 0-5 power supply of the Proto-Board provides a variable voltage. Use a 0MΩ ( 0 7 ohms), ¼ watt resistor for the series resistor SS. Measure SS with an ohmmeter (the DMM on the Ω scale) before building the circuit, since the actual resistance of a resistor may differ significantly from its nominal value. Your fingers are conductive, and can affect your resistance measurement, so keep them away from any metal parts in the circuit while measuring SS. ary the supply voltage between 0 and 5 in steps of about 3 (to get a voltage of 0 just turn off the power supply), and calculate the internal (input) resistance of DMM2 from these data using your calculation above (Use a fixed voltage range for DMM2 to make this measurement. Do not use autorange.). 0M S 2 DMM2 ideal voltmeter infinite DMM 0-5 volts 2 v s = 0 0 Figure 2: Circuit for measuring the internal resistance of voltmeter DMM2. 2
Figure 3: Photo of the setup for the internal resistance measurement in part and Figure 2. You can easily remove the large breadboard to see how the nodes and busses connect (see also Figure 4). 2 Internal esistance of an Ampère Meter To measure the current that flows through an electrical circuit, you can interrupt the circuit Figure 4: Details of Breadboard on Protoboard. Left, Top iew. ight, Wire connections between holes in board. Note that the holes at the top and bottom are connected together with long horizontal wires. The holes in the middle are connected in groups of five by short vertical wires. and insert an Ampère meter or Ammeter (a DMM set to a Current range.) so the current flows through the Ammeter. The resistance of an ideal ammeter is zero and does not change the current. However, a real ammeter has a small internal resistance AA, as in Figure 5. To 3
measure AA, set up the circuit shown in Figure 5. Use two kω ¼ watt resistors in parallel for SS. Make sure you measure SS before completing the circuit. Use the 5 volt variable supply in the Proto Board to vary the current into DMM2. Configure DMM2 as a current meter with range 0-20 ma and be sure to use the correct plugs on DMM2. Use DMM to measure the voltage across the real ammeter DMM2. Make sure you have DMM connected to measure, not SS. The connection is not the same as that in Part. If you measure SS rather than, you can still derive AA, but your calculation will be different and your result less accurate. Assuming DMM2 accurately reports the current, calculate AA =, where I is the current through the II ammeter. Note that by measuring and I you do not need to know the value of SS. If you measured SS and I, you would need to know SS very accurately to derive AA. The resistors will get warm when they dissipate a power near or above their rating of ¼W. Assuming that 0, calculate the voltage SS where the power dissipation in each resistor exceeds ¼ watt (remember PP = II 2 ). You can touch the resistors to sense this effect (be careful not to burn yourself). DMM2 (current meter) DMM S 0-5 volts k ea. S A A = 0 0 ideal current meter = 0 Figure 5: Circuit for measuring the internal resistance AA of ammeter DMM2. 3 Batteries as a oltage Source, Internal esistance of a Battery A battery is a device that produces a voltage difference between its poles. When we connect a load resistor LL between these poles, a current II flows through the load resistor. As we decrease L, the current through the resistor increases. However, the battery cannot supply very large currents, so for small LL, the voltage across the battery terminals drops. That is, the battery behaves like an equivalent circuit with an output resistance SS in series with 4
an ideal voltage supply. This output resistance limits the current flow the battery can produce. In order to measure SS, connect a variable load LL (resistor box) and a voltmeter across the battery as shown in Figure 6. You will obtain better results using a fresh battery. Use the same DMM2 and settings as in the previous part of the experiment to measure the current II through the load resistor. If LL is very large, II 0 and = SS. Assuming that the internal resistance of the voltmeter DMM is infinite, show that = SS ( LL+ AA ) LL + AA + SS. where A is the value you measured in part 2. Measure and I for different values of LL. Start with LL 000Ω and decrease the value of LL until the current increases to II 30mA. Using a very small L causes the battery to produce a very large current, which can damage the battery (we have plenty so don t worry too much). Use values of L which produce II > 30mA only briefly (for the duration of your measurement). Measure the actual values of LL that you used after you complete your measurements of and II. Measure SS before and after the experiment and use the average of the two values in your derivations. Make sure you include the difference between the before and after values of SS as a source of error in your error analysis. You can derive the value of SS in two different ways in this experiment. First consider the current in the entire circuit II. The voltage across DMM is SS minus the voltage across SS, = SS SS II. So: SS = SS. (eq. ) II In this case, you don t need to know AA or LL to measure SS. Alternatively, you know that: Substituting for II, II( AA + LL ) =, II = AA + LL. SS = SS ( AA + LL ). (eq. 2) In this case, we don t need to know II, which means we could have simplified the circuit by getting rid of the ammeter entirely (in which case the result would be SS = SS LL ). Calculate SS using both equation and equation 2. Are your results the same? Which is more accurate? Why? 5
Figure 6: Circuit for measuring the internal resistance SS of a battery. 4 Adjustable Power Supply and Output esistance of a oltage Divider Just as a battery has an internal resistance, a power supply also has an internal resistance, which means that the actual output voltage will be smaller than the nominal output voltage for loads with low resistance. We can explore the significance of this effect more easily if we build a power supply with an internal resistance we control. To explore the idea of an equivalent circuit in more detail, we will explore the use of a voltage divider to obtain a selected voltage from a fixed-voltage power supply. In Figure 7 we can think of the voltage across 2 as the output of a power supply producing a nominal voltage PPPP, (the voltage the equivalent power supply produces with no load ( LL = ) consisting of the real power supply plus the voltage divider provided by and 2 ). 6
Calculate 2 as a function of LL. Your result should show that the power supply plus divider behaves just like an equivalent power supply composed of an ideal voltage supply with a series resistance S, which turns out to be the parallel combination of resistors and 2 SS = 2. This result is an example of Thevenin's Theorem for two-terminal + 2 linear networks. If you are unfamiliar with Thevenin's Theorem, many web resources (such Equivalent Power Supply 2 DMM 5 volts s = 0 2 L DMM2 resistor box 00k... 500 ohms Figure 7: Circuit for creating an equivalent power supply with a selected voltage output using a fixed power supply and two resistors as a voltage divider. Equivalent Power Supply Thevenin's theorm v Equivalent Power Supply S 0 equivalent to 2 S PPPP PPPP Figure 8: Thevinin s theorem for analyzing the circuit in Figure 7 as an equivalent circuit, a power supply with nominal output voltage PPPP. The dashed boxes are equivalent circuits to the dashed box in Figure 7. as Wikipedia) explain it. If LL 2, then 2 PPPP. When LL 2 you will find 2 < PPPP. If LL 2 you will find 2 PPPP. Why? 7
To construct the circuit in Figure 7, use 0kkΩ and 5kkΩ for and 2. ary the resistance LL, using nominal values of LL = 0.5,.0, 3.3, 6.8, 0, 33, 00kΩ, (to get LL = just disconnect one terminal of the load resistor). Measure the actual values of these resistors with an ohmmeter before constructing the circuit. ecord the output voltage for each measured value of LL. Determine the expected value of SS in terms of SS, SS and SS. Compare your measured voltages with the expected values of SS. How well do your data match the theory? Questions to discuss in your discussion and conclusion sections: ) Questions related to part. Under what circumstances will the internal resistance of a voltmeter significantly affect the voltage you are trying to measure? Specifically, consider a situation in which the load resistance in the circuit between the two points at which you are measuring is, the remainder of the circuit has a series resistance 2 and the internal resistance of the voltmeter is. What will be the fractional difference mmmmmmmmmmmmmmmm uuuuuuuuuuuuuuuuuuuuuu in the voltage across between uuuuuuuuuuuuuuuuuuuuuu the unperturbed circuit and the circuit with the voltmeter in place as a function of, 2 and? Does this difference depend on uuuuuuuuuuuuuuuuuuuuuu? Define what you mean by Significant. Use your measured value of to determine a condition on and 2 so that the change in voltage is less than 5%. 2) Questions related to part 2. Under what circumstances will the internal resistance of an ammeter significantly affect the current you are trying to measure? Specifically, consider a situation in which the series resistance in the circuit you are measuring is and the internal resistance of the ammeter is AA. What will be the fractional difference II mmmmmmmmmmmmeedd II uuuuuuuuuuuuuuuuuuuuuu in the current through the circuit II uuuuuuuuuuuuuuuuuuuuuu between the unperturbed circuit and the circuit with the ammeter in place as a function of and AA? Does this difference depend on II uuuuuuuuuuuuuuuuuuuuuu? Define what you mean by Significant. Use your measured value of AA to determine a condition on so that the change in voltage is less than 5%. 3) Questions related to parts 3 and 4. a) Under what circumstances will the internal resistance of a battery or power supply significantly affect the voltage or current you are trying to apply to a circuit? Specifically, consider a situation in which the load resistance in the circuit you are measuring is LL and the internal resistance of the battery or power supply is SS. What will be the fractional difference aaaaaaaaaaaaaa uuuuuuuuuuuuuuuuuuuuuu in the uuuuuuuuuuuuuuuuuuuuuu voltage applied to the load as a function of LL and SS? Does this difference depend on uuuuuuuuuuuuuuuuuuuuuu? Define what you mean by Significant. Use your measured values of SS to determine a condition on LL so that the change in voltage is less than 5%. b) The current is a slightly different calculation. Imagine a perfect power supply which produced a current through your load II uunnnnnnnnnnnnnnnnnnnn = uuuuuuuuuuuuuuuuuuuuuu. Let LL II pppppppppppppppppp be the current through the real circuit. What is the fractional 8
difference in current II pppppppppppppppppp II uuuuuuuuuuuuuuuuuuuuuu II uuuuuuuuuuuuuuuuuuuuuu as a function of LL and SS? Is the dependence of the difference on LL and SS the same for current and voltage? Use your measured values of SS to determine a condition on LL so that the change in current is less than 5%. c) If you are designing a power supply to have a very well defined voltage output, should S be large or small? Why? If you are designing a power supply to have a very well defined current output should S be large or small? Why? Which is a better voltage supply, the battery, or the equivalent power supply in npart 4? Why? Which is a better current supply? Why? 9